POSITION, DISPLACEMENT

Kinematics is a branch of physics that is concerned with describing motion. The first concept of motion is position.

Think of your position at this instant. Another way of asking this question is “Where are you?”

Position is defined as the distance and direction from a reference point. A reference point may be in fact moving (e.g. Earth). Reference points are arbitrary choices made by us.

Position is a vector quantity because it has size or magnitude and direction. Scalar quantities have only size or magnitude. The symbol for position is

d

Direction can be given above or below the horizon, within a 2-D system (NEWS), or within a 3-D system (on Star Trek).

After the examples on the next slide complete the position worksheet.

For this class use the 2-D map system.

Record direction relative to north or south.

30.2o (measure)

15.41 cm (measure)

15.41 cm [N59.8oE]

11.6o (measure)

6.52 cm (measure)

6.52 cm [S11.6oW]

r.p.

DISPLACEMENT

Displacement is defined as change in position.

12 ddd

Displacement is a vector with magnitude and direction. It is found in the difference between two position vectors. Take note that it is always final position minus initial position.

Displacement does not need a reference point. Displacement remains the same even if the reference point was changed.

25.31 cm (measure)

r.p.

24.5o (measure)25.31 cm [S65.5oE]

initial point

final point

Try worksheet

Arrow points to final position!

Distance is defined as the length of path traveled. It is a scalar quantity and its symbol is

d

Solving problems involving displacement and distance demands we use proper form.

Adam is 59 m [E] of the flagpole when he chases Nick after Nick stole his lunch money. Later Adam is standing 24 m [W] of the flag pole. What was Adam’s displacement during the chase?

Kinematics Problems

-only one-dimensional problems will be studied

-always establish a direction system for each question

East is positive

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59

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][83

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Emor

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Note that the path Adam took from start to finish had no effect on displacement.

The answer works out just as well with West is positive.

Flagpole is r.p.

Meisha walks 350 m [N] and ends up standing 75 m [S] of the school. What was her starting position?

South is positive

School is r.p.

1

12

2

75350

75

350

d

ddd

md

md

][4251 Smd

Questions

1. Q stands 38.5 m [W] of Bell. He walks 82.7 m [E]. What is Q’s final position?

2. Stacey finishes a canoe trip 15.6 km [S] of her starting position. Her final position is 18.9 km [S] of Palmer Rapids. What was Stacey’s initial position?

3. Shawna jumps from a plane that is 9562 m above sea level and she pulls her rip cord at 1271 m above sea level. What was her displacement during her free fall?

Answers

1. 44.2 m [E]

2. 3.3 km [S]

3. 8291 m [down]

Displacement

At this point displacement is defined as

12

dddor change in position. Another way to calculate displacement is to sum up individual displacements.

...321

ddddRresultant or net displacement

Distance is defined as the sum of individual path lengths regardless of direction.

...321

dddd

Kiah takes her dog for a walk from her house, she walks 300 m [E] followed by 200 m [W].

a) What is her resultant displacement?

b) What is her total distance?

a) 21

dddR][3001 Emd

][2002 Emd

][100

200300

Emd

d

R

R

East is positive

md

d

ddd

500

20030021

b)

with no vector above, this tells us it is path length

Jason begins a walk 50 m [E] of Future Shop. Later he is 75 m [W] of Future Shop. Finally Jason is in Future Shop.(Assume Jason traveled in a line)

a)What was Jason’s displacement for each stage of his walk? (125 m [W], 75 m [E])

b)What was Jason’s displacement for the walk? (50 m [W])

c) What distance did Jason travel on his walk? (200 m)

Velocity and Speed

ARE YOU MOVING?

CAN YOU FEEL MOVEMENT?

HOW DO YOU KNOW IF YOU ARE MOVING?

If you are standing on the equator, how fast are you moving?

How fast is Earth moving?

Work with two others to solve these problems.

Ask me questions if you need certain facts.

Velocity is defined as the change in position with respect to time. It is a vector quantity.

t

dv

a vector divided by a scalar yields a vector

Speed is defined as distance traveled with respect to time. It is a scalar quantity.

t

dv

1. Aaron runs 40 m [E] on a soccer field in 28 seconds as she hustles to get the ball. Calculate her average velocity. (1.428 m/s [E])

2. Karan jogs at 52 m/s [NE] on his daily run. Today he ran 14.7 km [NE]. How long did he jog for? (282.7 s)

3. Mr. Moors runs at 1.8 m/s for 22 seconds while Stacy chases him uttering profanities. What distance did Mr. Moors travel during the chase? (39.6 m)

4. Jordyn walks 300 m [N] for 38 s then 178 m [S] for 62 s then 53 m [N] for 8 s in an effort to find Jackie. Her final position is 109 m [S] of the office.

a)What was Jordyn’s average velocity? (1.620 m/s [N])

b) What was Jordyn’s average speed? (4.917 m/s)

c) What was Jordyn’s initial position? (284 m [S])

d) What was Jordyn’s average velocity for the first two legs of her trip? (1.12 m/s [N])

Acceleration

Acceleration is defined as change in velocity with respect to time.

t

vva

12

t

va

In everyday usage acceleration often seems to mean speed up while decelerate means to slow down however technically this is not correct.

Acceleration is a vector quantity therefore has magnitude and direction. In one-dimension acceleration can have a positive or negative value. The sign indicates the sign of the velocity change since the change in time is always positive. Physicists do not mention deceleration. Speeding up refers to increasing magnitude of velocity while slowing down means the opposite.

LET’S DEMONSTRATE!!!

A runner accelerates from rest at 1.8 m/s2 [W]. The final velocity of the runner is 11.5 m/s [W]. How much time elapsed during this acceleration?

t

vva

12

s

mv

vs

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Wis

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8.1

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st 389.6

1. Billy travels at 14 m/s [NE] when hit by a truck, 0.145 s later Billy is travelling 52 m/s [SW]. What was B’s average acceleration during this time?(455.2m/s2[SW])

2. Gina accelerates at 2.89 m/s2 [N] for 14 s, Gina’s initial velocity was 15 km/hr [N]. What is G’s final velocity?(44.63 m/s [N])

3. Jen accelerates at 2.5 m/s2 [E] from 12.36 m/s [W] to 15.98 m/s [E]. How much time elapsed for this acceleration?

(11.34 s)

4. A car accelerates at 2481 km/hr2 [S] for 10 min. Its final velocity is 26 m/s [S]. What was the initial velocity of the car?

(88.86 m/s [N])

Constant Acceleration Kinematic Equations

Motion with a constant acceleration is an important type of motion which bears further analysis. In this unit only constant acceleration problems will be studied. The kinematic equations about to be shown or developed are only for

constant acceleration!

The average velocity during a constant acceleration is found with this equation.

212

vv

vav

tvv

d

)2

(12

So at this point when studying constant acceleration we have two equations.

t

vva

12

tvv

d

)2

(12

1. Billy accelerates uniformly from 10 m/s [E] to 21.3 m/s [W] in 13.4 s. a)What was Billy’s displacement during this time? (75.7 m [W]) b) What was Billy’s acceleration? (2.336 m/s2 [W])

2. A car travels 52 m [S] during 3.5 s of constant acceleration. The final velocity of the car was 77.76 km/hr [S] a) What was the car’s initial velocity?

(8.114 m/s [S]) b) What was the cars acceleration?

(3.853 m/s2 [S])

Constant Acceleration Kinematic Equations

The two constant acceleration equations below lead to 3 other kinematic equations.

tvv

d

)2

(12

t

vva

12

21

12

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2

2

)2

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vvt

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vvt

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vvta

t

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11

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( 12

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tvd

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tvd

-try and show how the last two are made

1. A turtle attained a velocity of 20 cm/s [E] after accelerating at 5 cm/s2 [E] for 3.5 s. a) What was the displacement of the turtle during this acceleration? (39.38 cm [E]) b) What was the turtle’s initial velocity?

(2.5 cm/s [E])

2. A bike accelerated from rest to a velocity of 42 km/hr [S] over a displacement of 87 m [S]. What was the acceleration of the bike? (0.7825 m/s2 [S])

3. A sportscar brakes hard and accelerates at 4.51m/s2 [S] over a displacement of 48.5 m [N] and a time of 2.31 s. What was the initial velocity of the car? (26.20 m/s [N])

4. An object starts from rest and accelerates at 3.0 ms-2 for 4.0 s. Its velocity remains constant for 7.0 s and it finally comes to rest with uniform deceleration after another 5.0 s.a) What is the displacement of each stage.b) What is the average velocity of the whole trip?

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s 4.0 t s

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m 3.0 a

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velocity theis (thiss

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b)

av

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5. A turtle is moving with a constant acceleration along a straight ditch. He starts his stopwatch as he passes a fence post and notes that it takes him 10 s to reach a pine tree 10 m farther along the ditch. As he passes the pine tree his speed is 1.2 ms-1. How far was he from the fence post when he started from rest?

p. 28 3-8 10-12 extra (graphs) p. 36 1-9 p. 46 1-5,7

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ACCELERATION DUE TO GRAVITY

Objects that have a net force of gravity acting upon them experience an accel-eration due to gravity of 9.81 m/s2 [down].

g = 9.81 m/s2 [down]

In these situations the force of air resistance is ignored. (terminal velocity, coffee filter and motion sensor) This acceleration applies to objects going up or down (or both).

position

time

An object is thrown upwards and returns to the same height.

velocity

time

slope = - 9.81 m/s2

acceleration

time

- 9.81 m/s2

position

time

at max. height v = 0

these two t’s are equal

the velocity at this time is negative of the initial velocity

An object is thrown upwards with a velocity of 24.2 m/s.

a)What is the object’s maximum height?

b)How long does it take the object to return to its starting height?

0

81.9

2.24

positive is up a)

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2.24 b)

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Kinematics

Describing how things move is called kinematics and it has terms that are very specific. These terms include position, displacement, distance, velocity, speed and acceleration. Proper use of these terms first demands that you understand each term’s definition, symbol and any equations associated with it.

Graphing

Graphs are often used in kinematics to represent position,velocity or acceleration (all versus time). Other quantities can be calculated from analysis of these graphs.

Curve SketchingDave begins west of the school and walks with a constant velocity and he passes the school. ([E] is positive)

t

d

This means position on a graph. We will never graph distance vs. time.

t

v

This means velocity on a graph. We will never graph speed vs. time.

Tanita walks back to school with a constant velocity, when she reaches the school she begins running with a constant velocity.

t

d

t

v

Karan runs away from school at a constant rate for a time then slows down until he stops.

t

d

t

v

t

a

A ball is thrown upwards and is caught below where it began.

t

d

t

v

t

a

Sketch a position vs. time graph, a velocity vs. time graph and an acceleration vs. time graph for the following situations.

a) A car travels forwards with a constant velocity of 50 km/hr [N].b) A person walks backwards with a velocity of 4 m/s [N].c) A person walks at 3 m/s for a period of time then jogs 6 m/s for the same period of time.d) A person walks at 3 m/s for a period of time then jogs home at 6 m/s in the opposite direction.e) A ball is thrown up out of a person’s hand; it rises and then returns to the person’s hand.

i) Sketch the graph for the velocity of the ball when it is out of the person’s hand.ii) Sketch the graph for the velocity of the ball for the complete

situation.f) A hockey puck slides along the smooth ice and bounces straight off the boards.g) A hockey puck is shot at a goalie and the goalie catches it.h) A person who slows down and speeds up without changing direction.i) A person who slows down and speeds up who does change direction. 

DOUBLE OBJECT PROBLEMS

-set up a direction system for each object such that each object moves in a positive direction

-there will always be a link between the two objects so that one variable can be eliminated

-extra information allows another equation to be written

Rowan steps from a helicopter at a height of 307 m and plummets to the earth. Mr. Moors (2 m tall) throws Rowan a football (from directly below) with a velocity 40 m/s 1 s after Rowan steps out . Assume the ball is released 2 m above the ground.

At what time and height will Rowan and the football meet?

307 m

2 m

2

81.9

2

81.9

0

down is

2

2

1

2

1

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Rowan football

2

81.940

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up is

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linked arent displaceme and time

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More problems on Double Object Problems sheet